A voltaic cell employs the following redox reaction:
Sn^2+(aq)+ Mn(s)--> Sn(s) + Mn^2+(aq)
Calculate the cell potential at 25 under each of the following conditions.
a.)[Sn]= 1.51E-2 M [Mn]= 2.52M
b.)[Sn]= 2.52 [Mn]= 1.51E-2 M
3 answers
That is 25 degree celsius
Obviously [Sn] can't be anything other than 1.00. You must mean [Sn^+2]. Same for Mn.
The easiest way to handle this is to calculate the half cell potential for each half cell, then add them together.
The reduction potential equation is
Ehalfcell = Eo-(0.0592/n)*log(red/ox)
For Sn that is
Ehalfcell = Eo (look up the potential as a reduction potential), then - (0.0592/2)*log(1/0.0151) = ?? (Note: The 1.00 is the value for Sn solid and 0.0151 is the value for Sn^+2 from the problem.)
Do the same thing for the Mn couple (but do it as a reduction), then reverse the equation, change the sign of the potential, and add the oxidation half to the reduction half. Post your work if you get stuck.
The easiest way to handle this is to calculate the half cell potential for each half cell, then add them together.
The reduction potential equation is
Ehalfcell = Eo-(0.0592/n)*log(red/ox)
For Sn that is
Ehalfcell = Eo (look up the potential as a reduction potential), then - (0.0592/2)*log(1/0.0151) = ?? (Note: The 1.00 is the value for Sn solid and 0.0151 is the value for Sn^+2 from the problem.)
Do the same thing for the Mn couple (but do it as a reduction), then reverse the equation, change the sign of the potential, and add the oxidation half to the reduction half. Post your work if you get stuck.
Thank you
1 answer